The same restriction digestion experiment was repeated for EcoR1 on the same DNA molecule. However, the incubation period of the reaction was lengthened for an additional 2 hours due to some unforeseen circumstances. Illustrate below the results you would observe after gel electrophoresis and explain your findings.
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- The restriction endonuclease NotI recognizes the octanucleotide sequence GCGGCCGC. Calculate the expected number of NotI cleavage sites in the bacteriophage λgenome, a linear DNA duplex 48.5 kbp in length with a (G + C) content of 50%.In the Sanger (dideoxy) method for DNA sequencing, a small amount of a dideoxynucleoside triphosphate—say, ddCTP—is added to the sequencing reaction along with a larger amount of the corresponding dCTP. What resultwould be observed if the dCTP were omitted?Could you explain how you would know if you successfully induced transformation? Thank you Did you successfully demonstrate transformation occurred? Why or why not? (How do I know, what do I look for?) Explain the control used in this experiment. What was it, and why was it necessary? Based on the calculation in the question above, comment on the relative difficulty of achieving successful transformation of E. coli. (Please explain the relative difficulty)
- They have purified a 1100 bp HindIII restriction fragment that they plan to sequence. As a first step, they decided to construct a restriction map of the fragment for the enzymes EcoRI and SmaI. Below is shown an agarose gel of the appropriate digests. Draw a restriction map of the fragment and show the distances, in base pairs, between the HindIII, EcoRI, and SmaI sites. *The restriction enzymes KpnI and Acc65I recognize and cleave the same 6-bp sequence. However, the sticky end formed from KpnI cleavage cannot be ligated directly to the sticky end formed from Acc65I cleavage. Explain why.Kornberg and his colleagues incubated soluble extracts of E. coli with a mixture of dATP, dTTP, dGTP, and dCTP, all labeled with 32P in the alpha-phosphate group. After a time, the incubation mixture was treated with trichloroacetic acid, which precipitates the DNA but not the nucleotide precursors. The precipitate was collected, and the extent of precursor incorporation into DNA was determined from the amount of radioactivity present in the precipitate. (a) If any one of the four nucleotide precursors were omitted from the incubation mixture, would radioactivity be found in the precipitate? Explain within 2 sentences. (2) (b) Would 32P be incorporated into the DNA if only dTTP were labeled? Explain within 2 sentences. (2) (c) Would radioactivity be found in the precipitate if 32P labeled the β or γ phosphate rather than the α phosphate of the deoxyribonucleotides? Explain within 2 sentences. (2)
- • Kornberg and his colleagues incubated soluble extracts of E. coli with a mixture of dATP, dTTP, dGTP, and dCTP, all labeled with 32P in the α-phosphate group. After a time, the incubation mixture was treated with trichloroacetic acid, which precipitates the DNA but not the nucleotide precursors. The precipitate was collected, and the extent of precursor incorporation into DNA was determined from the amount of radioactivity present in the precipitate.a. If any one of the four nucleotide precursors were omitted from the incubation mixture, would radioactivity be found in the precipitate? Explain.b. Would 32P be incorporated into the DNA if only dTTP were labeled? Explain.c. Would radioactivity be found in the precipitate if 32P labeled the or phosphate rather than the phosphate of the deoxyribonucleotides? Explain.Consider the DNA segment with a sequence: 3'-TACGGTACGGGATTG-5'. If the given DNA sample was subjected to pyrosequencing, sketch the expected profile of the sequencing output. Assume that the sequential flooding of nucleotides follows the order G-C-A-T.With regard to the experiment described in Figure, The DNA extract was treated with DNase, RNase, or protease.Why was this done? (In other words, what were the researcherstrying to determine?)
- Table 21.3 describes the cleavage sites of five different restrictionenzymes. After these restriction enzymes have cleaved the DNA, four of them produce sticky ends that can hydrogen bond with complementary sticky ends, as shown in Figure 21.1. The efficiency of sticky ends binding together depends on the number of hydrogen bonds; more hydrogen bonds makes the ends “stickier” and more likely to stay attached. Rank these four restriction enzymes from Table 21.3 (from best to worst)with regard to the efficiency of their sticky ends binding to each other.The restriction enzymes Kpn I and Acc 65I recognize and cleave the same 6-bp sequence. However, the sticky end formed from Kpn I cleavage cannot be ligated directly to the sticky end formed from Acc 65I cleavage. Explain why.Kpn I and Acc 65I are restriction enzymes that identify and cleave the same 6-bp sequence. The sticky end created by Kpn I cleavage, on the other hand, cannot be directly ligated to the sticky end formed by Acc 65I cleavage. Please explain why.